Question 1158401: Find how many distinct numbers greater than 5000 and divisible by 3 can be formed from the digits 3,4,5,6 and 0, if each digit is used atmost once in a number?
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website! Find how many distinct numbers greater than 5000 and divisible by 3 can be
formed from the digits 3,4,5,6 and 0, if each digit is used atmost once in a
number?
For an integer to be divisible by 3, the sum of its digits must be divisible by
3.
Case 1. The number has 5 digits.
Then it will be divisible by 3 because 3+4+5+6+0=18, which is divisible by 3.
All 5 digit numbers are greater than 5000, so
We choose the 1st digit 4 ways. (The first digit cannot be 0)
We choose the 2nd digit 4 ways.
We choose the 3rd digit 3 ways.
We choose the 4th digit 2 ways.
We choose the 5th digit 1 way.
So there are 4∙4∙3∙2∙1 = 96 five-digit numbers.
Case 2. The number has 4-digits and does not contain 0 as a digit.
Then it must be an arrangement of 3,4,5,6. Any such arrangement is divisible by
3 because 3+4+5+6 = 18.
Since it must be greater than 5000 there are only 2 choices for its first digit,
5 and 6.
We can choose the first digit 2 ways.
We can choose the second digit 3 ways.
We can choose the third digit 2 ways.
We can choose the fourth digit 1 way.
So there are 2∙3∙2∙1 = 12 ways for case 1
Case 3. The number has 4-digits and contains 0 as a digit.
Then the other 3 digits besides 0, must have a sum which is divisible by 3,
since the 0 will not change the sum of the digits.
So the other three digits must be 3,4,5 or 4,5,6, because 3+4+5=12, which
is divisible by 3, and 4+5+6=15, which is also divisible by 3.
[Note: The only other two combinations of digits 3,4,6 and 3,5,6 do not have
sums divisible by 3.]
Sub-Case 3A: Its digits are 3,4,5,0
We can choose the 1st digit 1 way, (as 5).
We can choose the 2nd digit 3 ways.
We can choose the 3rd digit 2 ways.
We can choose the 4th digit 1 way.
That's 1∙3∙2∙1 = 6 ways.
Sub-Case 3B: Its digits are 4,5,6,0
We can choose the 1st digit 2 ways, (as 5 or 6).
We can choose the 2nd digit 3 ways.
We can choose the 3rd digit 2 ways.
We can choose the 4th digit 1 way.
That's 2∙3∙2∙1 = 12 ways.
Answer: Adding all numbers for all the cases and sub-cases:
96+12+6+12 = 126 integers.
Edwin
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Edwin's solution is good, as usual.
Here is another approach, to show you there is almost always more than one way to analyze a problem.
The sum of the five given digits is 18, which is divisible by 3. Since all 5-digit numbers are greater than 5000, every 5-digit number that can be formed with these digits satisfies the requirements.
The first digit can't be 0; so we can choose...
the first digit 4 ways;
the second digit 4 ways;
the third digit 3 ways;
the fourth digit 2 ways; and
the fifth digit 1 way.
Number of 5-digit numbers we can make: 4*4*3*2*1 = 96.
For 4-digit numbers, the first digit must be either 5 or 6 if the number is going to be greater than 5000.
4-digit numbers with first digit 6....
With first digit 6, for the sum of the digits to be divisible by 3, the digits 4 and 5 must both be used. So we must use both of those digits and choose one of the remaining two digits; then those last three digits can be arranged in 3!=6 different ways.
Number of 4-digit numbers with leading digit 6: 2*6 = 12.
4-digit numbers with first digit 5....
With first digit 5, for the sum of the digits to be divisible by 3, the digit 4 must also be used. So we use the 4, and we choose 2 of the remaining 3 digits. We can choose 2 of the remaining three digits is C(3,2)=3 ways; then those last three digits can be arranged in 3!=6 different ways.
Number of 4-digit numbers with leading digit 6: 3*6 = 18.
Total number of numbers we can form: 96+12+18 = 126.
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